OFFSET
1,1
COMMENTS
This ratio is invariant of the size of the side-golden right triangle ABC. The shape of ABC is given by sidelengths a,b,c, where a=r*b, and c=sqrt(a^2+b^2), where r=(golden ratio)=(1+sqrt(5))/2. This is the unique right triangle matching the continued fraction [1,1,1,...] of r; i.e, under the side-partitioning procedure described in the 2007 reference, there is exactly 1 removable subtriangle at each stage. (This is analogous to the removal of 1 square at each stage of the partitioning of the golden rectangle as a collection of squares.)
Largest root of 4*x^4 - 20*x^2 - 20*x - 5. - Charles R Greathouse IV, May 07 2011
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..5000
Clark Kimberling, Two kinds of golden triangles, generalized to match continued fractions, Journal for Geometry and Graphics, 11 (2007) 165-171.
FORMULA
(circumradius)/(inradius)=abc(a+b+c)/(8*area^2), where area=area(ABC).
Equals (sqrt(5) + phi*sqrt(2 + phi))/2, where phi = A001622 is the golden ratio. - G. C. Greubel, Nov 23 2017
EXAMPLE
2.656875757337521549489732...
MATHEMATICA
r=(1+5^(1/2))/2; b=1; a=r*b; c=(a^2+b^2)^(1/2);
area = (1/4)((a+b+c)(b+c-a)(c+a-b)(a+b-c))^(1/2);
RealDigits[N[a*b*c*(a+b+c)/(8*area^2), 130]][[1]]
RealDigits[(Sqrt[5] + GoldenRatio*Sqrt[2 + GoldenRatio])/(2), 10, 50][[1]] (* G. C. Greubel, Nov 23 2017 *)
PROG
(PARI) {phi = (1 + sqrt(5))/2}; (sqrt(5) + phi*sqrt(2 + phi))/2 \\ G. C. Greubel, Nov 23 2017
(Magma) phi := (1+Sqrt(5))/2; [(Sqrt(5) + phi*Sqrt(2 + phi))/2]; // G. C. Greubel, Nov 23 2017
CROSSREFS
KEYWORD
AUTHOR
Clark Kimberling, Apr 05 2011
STATUS
approved